Question

If $f\left(x\right)=\sin \left(\log x\right)$ and $y=f\left[\left(\frac{2x+3}{3-2x}\right)\right],$ Find $\frac{dy}{dx}$ at $x=0$, if ans is in the form of $a/b$ in simplest form, find $a+b$

Answer 1

Since $f\left(x\right)=\sin \log x,$
we have $y=f\left(\frac{2x+3}{3-2x}\right)=\sin \log \left(\frac{2x+3}{3-2x}\right)=\sin \left[\log \right(2x+3)-\log (3-2x\left)\right]$
$\therefore \frac{dy}{dx}=\cos \left[\log \right(2x+3)-\log (3-2x\left)\right](\frac{2}{2x+3}+\frac{2}{3-2x})=\left(\frac{12}{9-4x^2}\right)\cos \log \left(\frac{2x+3}{3-2x}\right)$
$\therefore y^{\prime }\left(0\right)=\frac{4}{3}\Rightarrow a+b=7$